The one-point (Alexandroff) compactification $X^{\ast }=X\cup \{\mathrm{\infty }\}$, whose open sets are the open sets of $X$ together with the complements in $X^{\ast }$ of the closed compact subsets of $X$

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

A point outside $X$, named rather than assumed. Put

$\mathrm{\infty }:=\{y\in X:y\mathrm{\notin }y\},$

a set by Separation. Then $\mathrm{\infty }\mathrm{\notin }X$: were $\mathrm{\infty }\in X$, the defining condition applied to $\mathrm{\infty }$ itself would give $\mathrm{\infty }\in \mathrm{\infty }⟺\mathrm{\infty }\mathrm{\notin }\mathrm{\infty }$. So no hypothesis about $X$ is needed to obtain a point outside it, and the construction below is available for every space.

The space. Put $X^{\ast }:=X\cup \{\mathrm{\infty }\}$ and

${\mathcal{T}}^{\ast }:=\mathcal{T}\cup \{X^{\ast }\setminus C:C\subseteq X,C\text{ closed in }X\text{ and a compact subset of }X\}\mathrm{.}$

The pair $(X^{\ast },{\mathcal{T}}^{\ast })$ is the one-point compactification, or Alexandroff compactification, of $X$. Members of $\mathcal{T}$ are said to be of the first kind and the sets $X^{\ast }\setminus C$ of the second kind; a set of the second kind is exactly an open set of ${\mathcal{T}}^{\ast }$ containing $\mathrm{\infty }$, since a member of $\mathcal{T}$ is a subset of $X$, and the set $C$ is recovered from it as $C=X^{\ast }\setminus (X^{\ast }\setminus C)$.

${\mathcal{T}}^{\ast }$ is a topology on $X^{\ast }$, and this is discharged here. Throughout, "closed" and "compact" without qualification mean closed in $X$ and a compact subset of $X$ (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right); two facts about such sets are used and both are A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact: a subset of a compact $C$ that is closed in $X$ is closed in the subspace $C$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace) and hence compact, and a union of two compact subsets is compact.

(T1). $\mathrm{\varnothing }\in \mathcal{T}$, and $X^{\ast }=X^{\ast }\setminus \mathrm{\varnothing }$ is of the second kind, $\mathrm{\varnothing }$ being closed in $X$ and compact.

(T2). Let $\mathcal{S}\subseteq {\mathcal{T}}^{\ast }$, let ${\mathcal{S}}_1$ be the members of $\mathcal{S}$ lying in $\mathcal{T}$ and ${\mathcal{S}}_2$ the rest, so that every member of ${\mathcal{S}}_2$ is of the second kind. If ${\mathcal{S}}_2=\mathrm{\varnothing }$ then $\bigcup \mathcal{S}=\bigcup {\mathcal{S}}_1$ lies in $\mathcal{T}$ by (T2) in $X$. Otherwise put $U:=\bigcup {\mathcal{S}}_1\in \mathcal{T}$ and $\mathcal{D}:=\{X^{\ast }\setminus O:O\in {\mathcal{S}}_2\}$, a nonempty family of closed compact subsets of $X$, and $C_0:=\bigcap \mathcal{D}$. Then $C_0$ is closed by (C2) of Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, and it is a closed subset of any one member of $\mathcal{D}$, hence compact. Now

$\bigcup \mathcal{S}=U\cup (X^{\ast }\setminus C_0)=X^{\ast }\setminus (C_0\setminus U),$

and $C_0\setminus U=C_0\cap (X\setminus U)$ is closed in $X$ and a subset of the compact $C_0$, hence compact; so $\bigcup \mathcal{S}$ is of the second kind.

(T3). For $U,V\in \mathcal{T}$ the intersection lies in $\mathcal{T}$ by (T3) in $X$. For two sets of the second kind, $(X^{\ast }\setminus C)\cap (X^{\ast }\setminus D)=X^{\ast }\setminus (C\cup D)$, and $C\cup D$ is closed in $X$ and compact as a union of two compact subsets. For one of each, $\mathrm{\infty }\mathrm{\notin }U$ gives $U\cap (X^{\ast }\setminus C)=U\cap (X\setminus C)$, an intersection of two members of $\mathcal{T}$.

Why the compact sets are also required to be closed. The complement of a compact set that is not closed in $X$ would not make $\mathrm{\infty }$'s neighbourhoods behave: the union computation in (T2) uses that an intersection of the discarded sets is again closed, and the intersection of arbitrary compact subsets of a non-Hausdorff space need not be compact. When $X$ is Hausdorff every compact subset is closed (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones) and the two descriptions agree, which is why many texts state the definition without the word "closed" and silently assume the Hausdorff case.

What the name promises is proved, not assumed. That $X^{\ast }$ is compact, that $X$ sits inside it as an open subspace carrying its own topology, and the exact conditions under which $X$ is dense in $X^{\ast }$ or $X^{\ast }$ is Hausdorff, are $X^{\ast }$ is compact and contains $X$ as an open subspace; $X$ is dense in $X^{\ast }$ exactly when $X$ is not compact; and $X^{\ast }$ is Hausdorff exactly when $X$ is locally compact and Hausdorff ↗. Nothing above uses any of them.

The added point is a genuine construction and not a choice. The set $\mathrm{\infty }$ above is determined by $X$; no appeal to any principle of choice is made, and no "take a point not in $X$" is left unexplained.

$X^{\ast }$ is Hausdorff exactly when $X$ is locally compact and Hausdorff (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space), which is the reason local compactness and this construction always appear together.